= Solution
Keep $+x$ along the upper-sheet current and $+z$ upward, putting the lower sheet at $z=0$ and upper sheet at $z=d$. Between them, the upper sheet's lower-side <magnetic field> is $+\mu_0I/(2w)\widehat{\mathbf y}$. The lower sheet carries current in $-x$, so its upper-side <magnetic field> is also $+\mu_0I/(2w)\widehat{\mathbf y}$. The <superposition principle> therefore gives
$$
\boxed{\mathbf B=\frac{\mu_0I}{w}\widehat{\mathbf y}.}
$$
Each perfect conductor is an <equipotential>. The circuit's <resistor> carries current from upper to lower, so <Ohm's law> fixes their <electric potential difference>:
$$
V_{\rm upper}-V_{\rm lower}=IR.
$$
The broad parallel faces and their constant potentials make the <electric field> perpendicular to them and independent of the coordinates along them, away from edges. There is no volume <electric charge> in the gap. Apply <Gauss's law> to a small pillbox wholly within the gap, with equal face <areas> $A$ at heights $z_1,z_2$. Its side flux is zero, so
$$
[E_z(z_2)-E_z(z_1)]A=\frac{Q_{\rm enclosed}}{\varepsilon_0}=0.
$$
Thus $E_z$ is independent of height as well. Since $V_{\rm upper}-V_{\rm lower}=-\int_0^dE_z\,dz$,
$$
\boxed{\mathbf E=-\frac{IR}{d}\widehat{\mathbf z},\qquad E=\frac{IR}{d}.}
$$
It points from the higher-potential upper sheet to the lower sheet. <Gauss's law> requires the planar <symmetry> argument: zero enclosed charge alone would not prove a uniform <electric field> in a general geometry. These combined fields characterize <parallel current sheets with a resistive termination>.
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